From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: (ir)reducibility of polynomials
Date: 6 Oct 1998 16:58:34 GMT
Roberto Maria Avanzi wrote:
> Suppose f(x,y) is a polynomial in two variables x,y over the field
> K (can be the rationals, or a number field).
> Suppose further that for many, but finitely many, distinct values
> y_1,y_2,...,y_n of y the polynomial f(x,y_1) (which is now a polynomial in
> the solve variable x) is irreducible
> Then f(x,y) is irreducible (or the only factors you can factor
> out are of degree 0 in x, i.e. polynomials in the sole y).
Obviously if f(x,y) were reducible, say f(x,y) = g(x,y)*h(x,y), then
f(x,y_1) would be reducible for _every_ y_1, the factors being
g(x,y_1) and h(x,y_1). If your concern is that one or both of these
factors are of degree zero in x, you need only take any y_1 not in the
union of the zero loci of the coefficients of the highest powers of x
in each of g and h.
>Moreover I would like also to know if there is a (related ?) result
>like
>
> f(x,y) as above and with no factors which are polynomials in
> the sole y. if for many values y_i of y the polynomial f(x,y_i) is
> reducible, then the polynomial f(x,y) itself is reducible.
This is a little harder, but the Hilbert Irreducibility Theorem implies
that if f(x,y) were irreducible, then f(x,y_1) would be irreducible
in K[x] for almost all y_1 , if K is a subfield of the complex
numbers (using the usual topology on C).
I don't really know what algorithms are used by computer algebra systems
to factor multivariate polynomials, but in principle one could use this
idea: to factor f in K[x,y], factor f(x,y_i) completely for several
y_i, then consider all possible ways of choosing a factor g_i(x) of
f(x,y_i) and use Lagrange interpolation to construct a candidate factor
g(x,y) with g(x,y_i)=g_i(x) for all i; at some point you will either
succeed or know that this set of choices g_i is incompatible (since the
degree of g in y has exceeded the degree of f, for example).
dave